# ---
# title: 1170. Compare Strings by Frequency of the Smallest Character
# id: problem1170
# author: Tian Jun
# date: 2020-10-31
# difficulty: Easy
# categories: Array, String
# link: <https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/description/>
# hidden: true
# ---
# 
# Let's define a function `f(s)` over a non-empty string `s`, which calculates
# the frequency of the smallest character in `s`. For example, if `s = "dcce"`
# then `f(s) = 2` because the smallest character is `"c"` and its frequency is
# 2.
# 
# Now, given string arrays `queries` and `words`, return an integer array
# `answer`, where each `answer[i]` is the number of words such that
# `f(queries[i])` < `f(W)`, where `W` is a word in `words`.
# 
# 
# 
# **Example 1:**
# 
#     
#     
#     Input: queries = ["cbd"], words = ["zaaaz"]
#     Output: [1]
#     Explanation: On the first query we have f("cbd") = 1, f("zaaaz") = 3 so f("cbd") < f("zaaaz").
#     
# 
# **Example 2:**
# 
#     
#     
#     Input: queries = ["bbb","cc"], words = ["a","aa","aaa","aaaa"]
#     Output: [1,2]
#     Explanation: On the first query only f("bbb") < f("aaaa"). On the second query both f("aaa") and f("aaaa") are both > f("cc").
#     
# 
# 
# 
# **Constraints:**
# 
#   * `1 <= queries.length <= 2000`
#   * `1 <= words.length <= 2000`
#   * `1 <= queries[i].length, words[i].length <= 10`
#   * `queries[i][j]`, `words[i][j]` are English lowercase letters.
# 
# 
## @lc code=start
using LeetCode

## add your code here:
## @lc code=end
